In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $PR$. If $QR = 9$ and $PR = 13.5$,find $RM$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at point $D$,then

In $\Delta ABC$ and $\Delta DEF$,$\frac{AB}{DF} = \frac{BC}{EF}$ and $\angle B \cong \angle F$. Then,by $\ldots \ldots \ldots$ condition,the correspondence $ABC \leftrightarrow DFE$ is a similarity.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $CM = 3$,find $BM$.

$A$ and $B$ are points on the sides $PQ$ and $PR$ of a triangle $PQR$ respectively,such that $PQ = 12.5 \, cm$,$PA = 5 \, cm$,$BR = 6 \, cm$,and $PB = 4 \, cm$. Is $AB \parallel QR$? Give reasons for your answer.

In $\Delta ABC$,$D$,$E$ and $F$ are the midpoints of $\overline{BC}$,$\overline{CA}$ and $\overline{AB}$ respectively. Then,$\text{Area}(BDEF) = \ldots \times \text{Area}(\Delta ABC)$.